THE COMPARISON TESTS In the comparison tests the idea is to compare a given series with a series that is known to

11.4 THE COMPARISON TESTS In the comparison tests the idea is to compare a given series with a series that is known to

be convergent or divergent. For instance, the series

n苷 兺 1 2 n ⫹ 1

reminds us of the series 冘 ⬁ n苷 1 1 兾2 n , which is a geometric series with a苷 1 2 and r苷 1 2 and

is therefore convergent. Because the series (1) is so similar to a convergent series, we have the feeling that it too must be convergent. Indeed, it is. The inequality

shows that our given series (1) has smaller terms than those of the geometric series and therefore all its partial sums are also smaller than 1 (the sum of the geometric series). This means that its partial sums form a bounded increasing sequence, which is convergent. It also follows that the sum of the series is less than the sum of the geometric series:

n苷 1 ⫹ 1

Similar reasoning can be used to prove the following test, which applies only to series whose terms are positive. The first part says that if we have a series whose terms are smaller than those of a known convergent series, then our series is also convergent. The second part says that if we start with a series whose terms are larger than those of a known divergent series, then it too is divergent.

THE COMPARISON TEST Suppose that 冘 a n and 冘 b n are series with positive terms. (i) If 冘 b n is convergent and a n 艋b n for all , then n 冘 a n is also convergent. (ii) If 冘 b n is divergent and a n 艌b n for all , then n 冘 a n is also divergent.

N It is important to keep in mind the distinction

PROOF

between a sequence and a series. A sequence is

(i) Let

t苷

1 i苷 1 兺 b

a list of numbers, whereas a series is a sum. n

With every series 冘 a n there are associated two

i苷

n苷 1

Since both series have positive terms, the sequences 兵s n 其 and 兵t n 其 are increasing

sequences: the sequence 兵a n 其 of terms and the

sequence of 兵s n 其 partial sums.

共s n⫹ 1 苷 s n ⫹a n⫹ 1 艌s n 兲 . Also t n l t , so t n 艋t for all . Since n a i 艋b i , we have s n 艋t n . Thus s n 艋t for all . This means that n 兵s n 其 is increasing and bounded above and therefore

converges by the Monotonic Sequence Theorem. Thus 冘 a n converges. (ii) If 冘 b n is divergent, then t n l ⬁ (since 兵t n 其 is increasing). But a i 艌b i so s n 艌t n .

Thus . s n l ⬁ Therefore 冘 a n diverges.

In using the Comparison Test we must, of course, have some known series 冘 b n for

the purpose of comparison. Most of the time we use one of these series:

Standard Series for Use p N p A -series [ 冘 1 兾n converges if p⬎ 1 and diverges if p艋 1 ; see (11.3.1) ]

with the Comparison Test

A geometric series

[ n⫺ 冘 ar 1 converges if ⱍ r ⱍ ⬍1 and diverges if 艌 ⱍ r ⱍ 1 ;

see (11.2.4) ]

CHAPTER 11 INFINITE SEQUENCES AND SERIES

EXAMPLE 1 Determine whether the series 兺

converges or diverges.

2n ⫹ 4n ⫹ 3

n苷 1 2

SOLUTION For large the dominant term in the denominator is n 2n 2 so we compare the

given series with the series 冘 5 兾共2n 2 兲 . Observe that

2n 2 ⫹ 4n ⫹ 3

2n 2

because the left side has a bigger denominator. (In the notation of the Comparison Test,

a n is the left side and b n is the right side.) We know that

兺 2 苷 n苷 1 2n 2 n苷 兺 1 n 2

is convergent because it’s a constant times a -series with p p苷 2⬎1 . Therefore

n苷 兺 1 2n 2 ⫹ 4n ⫹ 3

is convergent by part (i) of the Comparison Test.

NOTE 1 Although the condition a n 艋b n or a n 艌b n in the Comparison Test is given for all , we need verify only that it holds for n n艌N , where N is some fixed integer, because the convergence of a series is not affected by a finite number of terms. This is illustrated in the next example.

n苷 兺 1 n

V EXAMPLE 2 ln n Test the series for convergence or divergence.

SOLUTION This series was tested (using the Integral Test) in Example 4 in Section 11.3, but it is also possible to test it by comparing it with the harmonic series. Observe that

We know that 冘 1 兾n is divergent ( -series with p p苷 1 ). Thus the given series is divergent

by the Comparison Test.

NOTE 2 The terms of the series being tested must be smaller than those of a convergent series or larger than those of a divergent series. If the terms are larger than the terms of a convergent series or smaller than those of a divergent series, then the Comparison Test doesn’t apply. Consider, for instance, the series

n苷 兺 2 1 n ⫺ 1

The inequality

is useless as far as the Comparison Test is concerned because n 冘 b n 苷 冘 1 ( 2 ) is convergent and a n ⬎b n . Nonetheless, we have the feeling that 冘 1 兾共2 n ⫺ 1 兲 n ought to be convergent because it is very similar to the convergent geometric series 冘 1 ( 2 ) . In such cases the fol-

lowing test can be used.

SECTION 11.4 THE COMPARISON TESTS

THE LIMIT COMPARISON TEST Suppose that 冘 a n and 冘 b n are series with positive

terms. If

N Exercises 40 and 41 deal with the

lim 苷 c

cases and . c苷

l 0 c苷⬁ n ⬁ b n

where c is a finite number and c⬎ 0 , then either both series converge or both

diverge.

PROOF Let m and M be positive numbers such that m ⬍c⬍M . Because a n 兾b n is close

to c for large n, there is an integer N such that

m⬍

⬍ M when n ⬎ N

and so

mb n ⬍a n ⬍ Mb n when n ⬎ N If 冘 b n converges, so does 冘 Mb n . Thus 冘 a n converges by part (i) of the Comparison

Test. If 冘 b n diverges, so does 冘 mb n and part (ii) of the Comparison Test shows that 冘 a n

diverges.